Question

The value of the integral \( \int_{0}^{1} \frac{x^3}{1 + x^8} \, dx \) is equal to:

• A. \( \frac{\pi}{8} \)
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{16} \)
• D. \( \frac{\pi}{6} \)
• E. \( \frac{\pi}{12} \)

Hint:

For integrals of the form $x^{n-1} / (1 + x^{2n})$, the substitution $u = x^n$ will always lead to an arctan result.

Answer 1

The Correct Option is C

Concept: We recognize that \( x^8 = (x^4)^2 \) and the derivative of \( x^4 \) is \( 4x^3 \). This suggests using the substitution \( u = x^4 \) to transform the integral into the standard \( \tan^{-1} \) form.

Step 1: Apply substitution.
Let \( u = x^4 \). Then \( du = 4x^3 \, dx \), so \( x^3 \, dx = \frac{1}{4} du \). Change the limits:
• When \( x = 0 \), \( u = 0 \).
• When \( x = 1 \), \( u = 1 \).

Step 2: Substitute and integrate.
\[ \int_{0}^{1} \frac{x^3}{1 + x^8} \, dx = \frac{1}{4} \int_{0}^{1} \frac{du}{1 + u^2} \] \[ = \frac{1}{4} [\tan^{-1} u]_{0}^{1} \]

Step 3: Evaluate.
\[ \frac{1}{4} (\tan^{-1} 1 - \tan^{-1} 0) = \frac{1}{4} \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{16} \]